Problem 27 - Calculus: Early Transcendentals

Question

Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty} \left(\sqrt{9x^2 + x} - 3x \right) \)

Step-by-Step Solution

Answer

The limit exists and is equal to \( \frac{1}{6} \).

Step 1: Simplify Expression in Limit

The expression under the limit is \( \sqrt{9x^2 + x} - 3x \). To simplify, rationalize the expression by multiplying both the numerator and the denominator by the conjugate, \( \sqrt{9x^2 + x} + 3x \). This results in:\[ \frac{(\sqrt{9x^2 + x} - 3x)(\sqrt{9x^2 + x} + 3x)}{\sqrt{9x^2 + x} + 3x} \]

Step 2: Multiply and Simplify

Expand the numerator: \((\sqrt{9x^2 + x})^2 - (3x)^2 = 9x^2 + x - 9x^2 = x\). Thus, the expression becomes:\[ \frac{x}{\sqrt{9x^2 + x} + 3x} \]

Step 3: Factor Out Highest Power of x in Denominator

The denominator can be simplified by factoring out \(x\): \[ \frac{x}{x(\sqrt{9 + \frac{1}{x}} + 3)} = \frac{1}{\sqrt{9 + \frac{1}{x}} + 3} \]

Step 4: Find the Limit as x Approaches Infinity

As \( x \to \infty \), the term \( \frac{1}{x} \to 0 \). Therefore, the expression becomes:\[ \lim_{x \to \infty} \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} \]

Step 5: Final Result

The limit is \( \frac{1}{6} \). Thus, the limit exists and is equal to \( \frac{1}{6} \).

Key Concepts

Rationalization

Rationalization is a technique used to eliminate radicals from the numerator or denominator of an expression. In this problem, we are dealing with the expression, \( \sqrt{9x^2 + x} - 3x \). Multiplying by the conjugate \( \sqrt{9x^2 + x} + 3x \) helps simplify the expression.

Why do we multiply by the conjugate? When you see a square root, especially when combined with other terms like subtraction or addition, think about multiplying by its conjugate. This turns the difference of squares into simpler forms, often eliminating square roots.

Here is how it works:

• The conjugate of \( a - b \) is \( a + b \).
• Multiplying the terms \( (a - b)(a + b) \) results in \( a^2 - b^2 \).
• Applying this technique, we simplify complex expressions under limits.

Infinity Limits

When finding limits as \( x \to \infty \), we assess behavior rather than calculating direct values. Infinity limits focus on understanding how functions behave for very large values of \( x \).

Here, we see that as \( x \to \infty \), the terms in the equation involving \( \frac{1}{x} \) shrink towards zero. Hence, the dominant terms become crucial.

• In limits involving infinity, compare powers of \( x \).
• Don't forget: \( \frac{1}{x} \to 0 \) as \( x \to \infty \).

These insights help isolate terms that genuinely affect the eventual limit.

Conjugates

Conjugates are instrumental when working with radicals, notably simplifying expressions in limits. Using the earlier exercise, the square root expression was paired with its conjugate:\( \sqrt{9x^2 + x} + 3x \).

This strategic move expands the expression using the difference of squares formula. It allows for straightforward algebraic manipulation:

• Identify the expression: here, \( \sqrt{9x^2 + x} - 3x \).
• Find the conjugate, which is \( \sqrt{9x^2 + x} + 3x \).
• Use the identity: \( (a - b)(a + b) = a^2 - b^2 \).

This approach efficiently removes the square root, leading to an easily solvable linear form.

Limit Simplification

Limit simplification helps to make complex limit problems solvable. Through steps like rationalization and working with conjugates, difficult expressions simplify incrementally. For instance, after deploying the conjugate method, the expression \( \frac{x}{\sqrt{9x^2 + x} + 3x} \) arises.

Taking factors out systematically simplifies further:

• In the denominator, factor out the highest power of \( x \).
• As \( x \to \infty \), focus on components independent of \( x \) for simplicity.
• Simplified form: \( \frac{1}{\sqrt{9 + \frac{1}{x}} + 3} \), highlights the dominant terms.

This is how we discover the limit to be \( \frac{1}{6} \), ensuring correctness and completeness in solving for limits.